Verify? $(cotx-tanx)/cosxsinx=csc^2x-sec^2x$

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Answer 1 · 2016-11-28T08:39:42

See below:

Let's start with the original:

$(cotx-tanx)/cosxsinx=csc^2x-sec^2x$

$(sinx/cosx)cotx-(sinx/cosx)tanx=csc^2x-sec^2x$

$(sinx/cosx)(cosx/sinx)-(sinx/cosx)(sinx/cosx)=csc^2x-sec^2x$

$1-(sin^2x/cos^2x)=csc^2x-sec^2x$

$1-tan^2x=csc^2x-sec^2x$

$color(red)("recall the identity" tan^2x=sec^2x-1$

$1-(sec^2x-1)=csc^2x-sec^2x$

$1-sec^2x+1=csc^2x-sec^2x$

$2-sec^2x!=csc^2x-sec^2x$

So as written the identity doesn't work. I suspect that what was meant was for the left hand $sinx$ term to be in the denominator. Let's try that:

$(cotx-tanx)/(cosxsinx)=csc^2x-sec^2x$

$cotx/(cosxsinx)-tanx/(cosxsinx)=csc^2x-sec^2x$

$(cosx/sinx)/(cosxsinx)-(sinx/cosx)/(cosxsinx)=csc^2x-sec^2x$

$cosx/(sinxcosxsinx)-sinx/(cosxcosxsinx)=csc^2x-sec^2x$

$cancelcosx/(sinxcancelcosxsinx)-cancelsinx/(cosxcosxcancelsinx)=csc^2x-sec^2x$

$1/(sin^2x)-1/(cos^2x)=csc^2x-sec^2x$

$csc^2x-sec^2x=csc^2x-sec^2x$

Verify? (cotx-tanx)/cosxsinx=csc^2x-sec^2x(cotx-tanx)/cosxsinx=csc^2x-sec^2x